Ramification filtration for automorphism group of Artin-Schreier curve I am studying the curve over the algebraic closure of $\mathbb{F}_3 = K$ defined by the equation $y^3 - y = x^4$. The automorphism group I am looking at is the one generated by elements $\sigma$ and $\tau$ where 
$\sigma^*(x) = x$, $\sigma^*(y) = y+1$
$\tau^*(x) = \zeta x$, $\tau^*(y) = \zeta^4 y$.
Here, $\zeta$ is a primitive 8-th root of unity and so we can also write $\tau^*(y) = -y$. One can easily check that $\tau \sigma = \sigma^2 \tau$, and so $G$ is isomorphic to $\mathbb{Z} / 3\mathbb{Z} \rtimes \mathbb{Z} / 8\mathbb{Z}$.
I now want to study the ramification over the point $\infty$, and in particular the lower ramification filtration. I then change affine charts and localize around 0 so that we end up looking at an extension of power series rings. The issue I run into is that $\tau^2$ acts as the identity (in both affine charts) on y, so the artin character is not defined for $\tau^2$. 
My question is this: Does it not make sense to look at the ramification filtration for this cover? 
 A: As Jyrki Lahtonen pointed out, we are considering a tower of fields given by $K(x^8) \subset K(x) \subset K(x,y)$. When we localize around $\infty$ we end up with a totally ramified extension $K((\tilde{x}, \tilde{y})) / K((\tilde{x}^8))$. I use the ~ to denote we've changed affine open sets, and the fact this is totally ramified follows $K$ being algebraically closed.
As $K(\tilde{x}) / K(\tilde{x^8})$ is Galois, we may consider the group $G/<\sigma> = G/H$ which is isomorphic to $\mathbb{Z}/8\mathbb{Z}$. Using the notation of Serre, we have that
$i_{G/H}(\tau) = \frac{1}{e'}\sum_{s->\sigma} i_G(s)$ where $e' = e_{K(\tilde{x}, \tilde{y})/K(\tilde{x})} = 3$. Just by considering the cardinalities of the groups, we have that $|\{ s\in G: s-> \tau\}|=3$.
Finally, for any $\tau \in G/H$, we have that $\tau(\tilde{x}) = x/a$ where $a\in K$ satisfies $a^8 = 1$. Thus $i_{G/H}(\tau^k) = 1$ for all $k \neq 8$.
It follows that $i_G(s) = 1$ for all $s\neq 1$. Certainly each $s\in G$ satisfies $i_G(s) \geq 1$ by the fact this is totally ramified, but it is at most one as $i_{G/H}(\tau) = 1 = \frac{1}{e'}\sum_{s->\sigma} i_G(s) = \frac{1}{3}\sum_{s->\sigma} i_G(s)$.
Thus the lower ramification filtration is $G = G_0 \supset \{1\}$.
